1. An air-core solenoid contains 1060 m1 turns of wire per meter and has a cross
ID: 1462184 • Letter: 1
Question
1. An air-core solenoid contains 1060 m1 turns of wire per meter and has a cross-sectional area of 0.159 m2 . A current of 3.2 A is established in the solenoid and is then reduced linearly to zero in 5.77 s. A pickup coil of 50 turns circles the solenoid near its center. What is the magnitude of the induced emf in the pickup coil? Answer in units of V. ANSWER = 5.87 X 10^-3
2. If the resistance of the loop is 0.00401 , what is the induced current? Answer in units of A.
I NEED THE ANSWER FOR QUESTION NUMBER 2. THE ANSWER FOR NUMBER 1 IS 5.87 X10-3
Explanation / Answer
1) magnetic field due to solenoid = B = uo*n*I
flux throught the pickup coil = phi = N*B*A
emf induced = rate of chnage in flux = (d/dt)* phi
emf = (d/dt)*(N*uo*n*I*A) = N*uo*n*A*(dI/dt)
dI = 3.2 A
dt = 5.77 s
A = 0.0159 m^2
emf = 50*4*pi*10^-7*1060*0.159*3.2/5.77
emf = 5.87*10^-3 V
2)
induced current = emf/R = 5.87*10^-3/0.00401 = 1.464 A
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